Question

For a first order process $A\to B$ rate constant $k_1=0.693{min}^{-1}$ and another first order process $C\to D$, $K_2=x$ ${min}^{-1}$. If $99.9$% of $C\to D$ requires time same as $50$% of reaction $A\to B$, value of $x$? (in ${min}^{-1}$)

• A. $0.0693$
• B. $6.93$
• C. $23.03$
• D. $13.86$

Answer 1

The correct option is D $13.86$

The solution is as follows:
1. Time required for $50$% completion for reaction $1$ is:

$t_{1/2}=\frac{0.693}{k}$

Given $k=0.639mi{n}^{-1}$

So $t_{1/2}=\frac{0.693}{0.693}=1min$

2. For reaction $2$, it takes $1$ min to get $99.99$% of reaction to be completed. Thus, it takes $0.5$ min to get $50$% reaction.
Thus,

$t_{1/2}=\frac{0.693}{K_2}$

$K_2=\frac{0.693}{t_{1/2}}$

$K_2=\frac{0.693}{0.5}$

Option D is correct.

$K_2=13.86mi{n}^{-1}$